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Question 22N.2.HL.TZ0.b.ii

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Date November 2022 Marks available [Maximum mark: 2] Reference code 22N.2.HL.TZ0.b.ii
Level HL Paper 2 Time zone TZ0
Command term Determine Question number b.ii Adapted from N/A
b.ii.
[Maximum mark: 2]
22N.2.HL.TZ0.b.ii

A satellite is launched from the surface of Earth into a circular orbit.

The following data are given.

                                        Mass of the satellite = 8.0 × 102 kg

Height of the orbit above the surface of Earth = 5.0 × 105 m

                                                 Mass of Earth = 6.0 × 1024 kg

                                              Radius of Earth = 6.4 × 106 m

Determine the minimum energy required to launch the satellite. Ignore the original kinetic energy of the satellite due to Earth’s rotation.

[2]

Markscheme

change in PE =6.67×10-11×6.0×1024×8.0×10216.4×106-16.9×106= «3.6×109 J» ✓

energy needed = KE + ΔPE = 2.7×1010 «J» ✓

 

Allow ECF from 8(b)(i).

Examiners report

Generally, this question was not well done. Most carried out a calculation based on the formula for escape velocity. An opportunity to remind candidates of reading back the stem for the sub-question when answering a second or any subsequent part of it.

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